Chapter 7: Numerical Series

Chapter 7: Numerical Series

Chapter 7 Overview: Sequeces ad Numerical Series I most texts, the topic of sequeces ad series appears, at first, to be a side topic. There are almost o derivatives or itegrals (which is what most studets thik all Calculus is comprised of) ad those that do appear do so at the ed of the chapter. The sese of discoectedess is heighteed by the fact that most Calculus studets have ot see series sice freshma year (ad, the, oly briefly). For the purposes of AP, this topic is broke ito four basic subtopics: Numerical sequeces ad series Radius ad iterval of covergece for a Power Series Taylor polyomials Creatig a ew series from a old oe This is the order most texts use, ad all topics are i oe chapter, which ca be overwhelmig. The fial subtopic is where the derivatives ad itegrals reappear, built o the theoretical foudatios laid by the first three. We are optig to break the topic of sequeces ad series ito two chapters. This chapter covers covergece ad divergece of umerical sequeces ad series. The other chapter will cover the material related to Power Series ad occurs at the ed of the curriculum. What we will cosider i Calculus is the issue of whether a sequece or series coverges or ot. There are several processes to test if a series is coverget or ot. There are seve Tests to prove covergece or divergece of a umerical series.. The Divergece (th Term) Test 2. The Compariso Test 3. The Limit Compariso Test 4. The Itegral Test 5. The Ratio Test 6. The th Root Test 7. The Alteratig Series Test Each may be easier or more difficult depedig o the series beig tested. Obviously, we will ot cover #4 i this chapter, sice we have ot leared Itegratio yet. The rest rely o Limits-at-Ifiity, ad they provide a good 448

cotext withi which to practice that material. Most of these Tests apply to series comprised of positive values. The Alteratig Series Test applies to series i which the terms alterate sigs. Though each sectio itroduces a ew test, the homework i each sectio will apply to all the tests that were itroduced previously. 449

7.: Sequeces Algebra Review Let s have a quick review from Algebra 2: A sequece is simply a list of umbers. Sequeces are oly iterestig to a mathematicia whe there is a patter withi the umbers. For example:, 3, 5, 7, 4, 7, 0, -7, -4,, 3, 9, 27, 4, 2,, 2, 4,... I Algebra, a sequece is writte as,, 2, 3, 5, 8, 3, a, a 2, a 3, a 4, a 5,..., a,... The geeral terms i the sequece have a subscript idicatig the cardiality--that is, which place they are i the sequece. a is the first umber i the sequece, a 2 is the secod, ad so o. a is the th umber. Note that was always a coutig umber. I calculus ad physics, this will chage i that the subscript may be tied to the expoet ad the series will geerally be writte a 0, a, a 2, a 3, a 4,..., a,... Vocabulary Sequece-- a fuctio with the Natural Numbers as the domai. Arithmetic Sequece-- A sequece i which oe term equals a costat added to the precedig term, i.e. a + = a + d. 450

Geometric Sequece-- A sequece i which oe term equals a costat multiplied to the precedig term, i.e. a + = a r. Series--the sum of a sequece. Partial Sum--Def: the sum of the first terms of a sequece. Much of what we did with sequeces i Algebra 2H ivolved fidig a (the th term i the sequece) or S (the sum of the first umbers i the sequece) directly without havig to fid all the umbers i the sequece. The patter of the sequece leads to a equatio where a is determied by, ad that patter is determied by the type of sequece it is. It was basically a cotext for arithmetic. OBJECTIVES Idetify Sequeces ad Series Fid Partial Sums of a give Series. Fid the terms, partial sums, ifiite sums, or i a geometric sequece. 45

Ex Fid the ext three terms ad the geeral term i each of these sequeces ad classify them as a arithmetic sequece, a geometric sequece or either. a), 3, 5, 7, b) 4, 7, 0, -7, -4, 4, 2,, 2, 4,... c) d),, 2, 3, 5, 8, 3, e), (f) 4, 9, 6, 25,..., 2, 3, 4, 5,... a) 9,, 3,, + 2( ). It is arithmetic because we add 2 each time. b) -2, -28, -35,, 4 7( ). It is arithmetic because we add -7 each time. c). It is geometric because we multiply by 8, 6, 32,..., 2 2 each time. 2 d) 2, 34, 55. This is either arithmetic or geometric. We do ot multiply by or add the same amout each step. It is called the Fiboacci Sequece. e). This is a p-series with p=2. 36, 49, 64,..., 2 + 6, 7,,..., 8 f). This is the Alteratig Harmoic Series. Sequeces are geerally writte Set Notatio: { a } Series are geerally writte with sigma otatio: a k k= 452

where is the k th term i the sequece ad we are substitutig the itegers a k through i for k, gettig the terms ad addig them up. meas sum. There is a symbol for the product of a sequece ( Π), but we will ot go ito that here. The first two examples at the begiig of the sectio are arithmetic sequeces. I, 3, 5, 7,, the differece (d) betwee ay two cosecutive terms is 2. I 4, 7, 0, -7, -4,, the differece is -7. Sice the differece is costat, we ca jump from the first term to ay term i the sequece by addig oe less d tha the umber of terms. The fifth term is the first plus four d s (or plus 4d). The fifth term is four steps away from the first. Arithmetic Sequece a = a + ( )d Ex 2 Fid a 32 for a arithmetic sequece with a = 6 ad d = 3 a = a + ( )d a 32 = 6 + ( 32 )3 = 99 The arithmetic sequece formula a = a + ( )d clearly has four variables i it: a, a,, ad d. Give ay three, we ca fid the fourth. Ex 3 Fid the first term i the arithmetic sequece where term is -65. d = 4 ad the 5 th 3 453

a = a + ( )d 65 = a + ( 5 ) 65 = a + 56 3 a = 46 3 4 3 Ex 4 If 2 is a term i the arithmetic sequece with term it is. a = 5 ad d = -7, fid which a = a + ( )d 2 = 5+ So 2 is the eighth term i the sequece. ( 7) 2 = 5 7 + 7 7 = 56 = 8 Note tha we would also kow if 2 was ot i the sequece by the ature of the i our solutio. If had ot bee a whole umber, 2 would ot be i the sequece. We ca fid several terms i a sequece if we kow the first ad last ad how may there are. These terms are sometimes called the meas because they are evely spaced betwee the kow values. Ex 5 Fid six arithmetic meas betwee 27 ad 44. This meas the sequece is 27,,,,,,, 44, or a = 27 ad a 8 = 44 This gives us eough iformatio to fid d: 454

a = a + ( )d 44 = 27 + ( 8 )d 7 = 7d d = 7 7 = 2 3 7 d, of course, ca be a fractio. The sequece, the, must be 27, 29 3 7, 36 7, 34 2 7, 36 5 7, 39 7, 44 7, 44 The third ad fourth examples at the start of this sectio are geometric sequeces. I, 3, 9, 27,, the ratio (r) betwee ay two cosecutive terms is 3. I 4, 2,,, the ratio is. 2, 4,... 2 Sice the ratio is costat, we ca jump from the first term to ay term i the sequece by multiplyig oe less r tha the umber of terms. The fifth term is the first times r four times (or ). r 4 Geometric Sequece a = a r As with the arithmetic sequeces, the geometric sequece formula a = a r clearly has four variables i it: a, a,, ad r. Give ay three, we ca fid the fourth. Ex 6 Fid a 5 for a geometric sequece with a = 024 ad r = 2 455

a = a r a 5 = 024 2 = 6 Ex 7 Fid the first term i the geometric sequece where is 2869784. 4 a = a r 2869784 = a ( 3) 4 a = 6 r = 3 ad the 5 th term Ex 8 If 2 is a term i the geometric sequece with a = 5 ad r =, fid which 2 term it is. a = a r 2 = 5 2 2 5 = 2 - Solvig for i the expoet will require a logarithm. - 456

l 2 5 = l 2 - l 2 5 = ( )l 2 = l 2 5 l 2 = l 2 5 l 2 = 5.672 + Sice this is ot a iteger, 2 is ot i the sequece. Ex 9 Fid four geometric meas betwee ad 325. This meas the sequece is,,,,, 325, or a = ad a 6 = 325 This gives us eough iformatio to fid r: a = a r 325 = r 5 5 r = 325 = 5 r, of course, ca be a fractio. The sequece, the, must be, 5, 25, 25, 625, 325 457

7. Homework. Fid the 2 th term i 2, 5, 8, 2. Fid the 49 th term i 2, -6, -4, 3. Fid a 22 for a arithmetic sequece with a = 35 ad d = -5. 4. Fid for a arithmetic sequece with a = 4 ad d =.5. a 39 5. Fid the first term i the arithmetic sequece where d = 2 ad the th term 3 is 243. 6. Fid the first term i the arithmetic sequece where d = 2 ad the 3st term is 45. 7. If 2 is a term i the arithmetic sequece with a = 42 ad d = -6, fid which term it is. 8. If 20 is a term i the arithmetic sequece with a = 22 ad d = 7, fid which term it is. 9. Fid three arithmetic meas betwee 42 ad 70. 0. Fid five arithmetic meas betwee -07 ad -86.. Fid six arithmetic meas betwee 47 ad -. Determie whether the sequece is arithmetic, geometric, or either. If arithmetic or geometric, defie d or r. 2. 3. 4. 5. 8, 3, 8, 23,..., 3, 9, 27,... 25, 50, 00, 200,... 25, 75, 00, 25,... 458

6. 7., 2, 3, 4,... 3 6 5, 5, 5,,,... 6 5 Fid the desigated variable. 8. Fid the 2 th term i 2, 6, 8, 9. Fid the 49 th term i, -2, 4, 20. Fid a 37 for a geometric sequece with a = 9 ad r =.88. 2. Fid a 6 for a geometric sequece with a = ad r = 2. 3 22. Fid the first term i the geometric sequece where r =.95 ad the 3 st term is 24.64. 23. Fid the first term i the geometric sequece where r = 2 ad the 4th term 3 is 243. 32 24. If 536 is a term i the geometric sequece with a = 3 ad r = 2, fid which term it is. 25. If is a term i the geometric sequece with a = 729 ad r =, fid which 3 term it is. 26. Fid two geometric meas betwee 5 ad 35. 27. Fid three geometric meas betwee ad 25. 525 2 28. Fid three geometric meas betwee 5 ad 45. 459

7.2 Series Vocabulary Series--the sum of a sequece. Partial Sum--Def: the sum of the first terms of a sequece. Ex Fid 4 k=0 ( k 2 ) 4 ( k 2 ) = ( 0 2 )+ ( 2 )+ ( 2 2 )+ ( 3 2 )+ 4 2 k=0 =+ 0 3 8 5 = 25 Ex 2 Fid 5 k=2 3 k 2 5 3 = 3 k=2 k 2 5 k=2 k 2 = 3 2 + 2 3 + 2 4 + 2 5 2 = 3 4 + 9 + 6 + 25 =.39 I a arithmetic sequece, the sum of the first ad last terms is the same as the sum of the secod ad secod to last, which is the same as the sum of the third ad third to last. There are half as may pairs of umbers i a sequece as there are umbers i a sequece, so the partial sum is S = ( 2 a + a ) 460

Sice we have a formula a = a + ( )d to get ay term i the sequece from the first term, we ca create a formula for the sum of the first terms. Arithmetic Partial Sum S = 2 ( 2a + ( )d) Ex 3 Fid S 32 for a arithmetic sequece with a = 6 ad d = 3 As with the arithmetic sequece formula a = a + ( )d, S = 2 2a + ( )d has four variables i it: ( + ( 3)3) S = 2 2a + ( )d S 32 = 32 2 2 6 =680 S, a,, ad d. Give ay three, we ca fid the fourth. Ex 4 Fid the first term i the arithmetic series where sum is -65. d = 4 3 ad the 3 th partial S = 2 ( 2a + ( )d) 65 = 3 2 2a + ( 3 ) 0 = 2a 6 a = 3 4 3 46

Ex 5 30705 is a partial sum of the arithmetic series with a = 7 ad d = 3. Which sum is it? ( + ( ) ( 3) ) S = 2 2a + ( )d 30705 = 2 2 7 0 = 3 2 + 3 640 640 = 34 + 3 3 = 3± 32 4 3 6 =38 or 48 3 ( 640) Sice must be a iteger, = 38 462

Just as with the arithmetic series, there is a formula for the th partial sum of a geometric series. Geometric Partial Sum S = a r r As with the arithmetic sequeces, the geometric sequece formula a = a r clearly has four variables i it: S, a,, r. Give ay three, we ca fid the fourth. Ex 6 Fid S 5 for the geometric sequece with a = 024 ad r = 2 S = a r r 2 S 5 = 024 2 = 682.625 4 Ex 7 Fid the first term i the geometric sequece where sum is 24605. r = 3 ad the 9 th partial S = a r r 9 3 24605 = a 3 a = 5 463

Ex 8 If S = 026, a = 6 ad r = 2, fid. r S = a r 2 2 026 = ( 6) 53 = 2 52 = 2 Sice we kow we caot log a egative, the egatives i this equatio must cacel, ad 52 = 2 52 = 2 L 52 = L 2 L 52 = L2 = 9 464

Sum of a Ifiite Geometric Series S = a r if ad oly if r <., Ex 9 Fid the ifiite sum for the geometric sequece with a = 024 ad r =. 2 S = a r S = 024 2 = 682 2 3 Ex 0 Fid the ifiite sum for the geometric sequece with a = 24 ad r = 5. 4 Sice r = 5, the formula will yield a icorrect aswer. This series is 4 > diverget. 465

7.2 Homework Fid the sum. 5 4. 2. 3k 5 k k= k= 5 k ( 2k +) 3. 4. k= 7 2 k k=3 5 5. 6. k=0 k 2 3 k+2 7 k=3 ( 2 + k 2 ) Fid the desigated variable. 7. Fid S 0 for a arithmetic series with a = 35 ad d = -5. 8. Fid for a arithmetic series with a = 4 ad d =.5. S 5 9. Fid S 3 for a arithmetic series with a 2 = 4 ad d = -3. 0. Fid i the arithmetic series where a = 5, d = 4 ad S = 329.. Fid i the arithmetic series where a = 8, d = 5 ad S = 4859. 2. Fid the first terms i the arithmetic series where d = 3 ad S 0 = 75. 3. Fid the first term i the arithmetic series where d = 2 ad S 5 = 0. 4. Fid the sum of the first 2 terms of 2, 5, 8, 5. Fid the sum of the first 60 terms of 2, -6, -4, 6. Fid the sum of the first 2 terms of 2, 6, 8, 7. Fid the sum of the first 60 terms of, -2, 4, 466

8. I the geometric sequece from a = 6 to a = 96608 where r = -2, how may terms are there ad what is the partial sum? 9. Fid the sum of the ifiite geometric series where a = 5 ad r =.4. 20. Fid the sum of the ifiite geometric series where a = 8 ad r = 3. 7 2. Fid the sum of the ifiite geometric series where a = 42 ad r = 2. 467

7.3: Covergece ad Divergece of Ifiite Series OBJECTIVES Determie the covergece or divergece of a sequece. Determie the divergece of a series. Use the Alteratig Series Test to check for covergece or divergece. Most of the covergece tests will ivolve fidig a limit at ifiity. It would be worth reviewig this topic from the Limits Chapter. Remember: Coverget Itegral--Def: a improper itegral that has a total. Diverget Itegral --Def: a improper itegral that does ot have a total. lim ta x = π x 2 or lim ta x = π x 2 x ± lim x = 0 lim e x = ad x lim e x = 0 x 468

The Hierarchy of Fuctios. Logs grow the slowest. 2. Polyomials, Ratioals ad Radicals grow faster tha logs ad the degree of the Ed Behavior Model (EBM) determies which algebraic fuctio grows fastest. For example, y = x 2 grows more slowly tha y = x 2. 3. The trig iverses fall i betwee the algebraic fuctios at the value of their respective horizotal asymptotes. 4. Expoetial fuctios grow faster tha the others. (I BC Calculus, we will see the factorial fuctio, y =!, grows the fastest.) 5. The fastest growig fuctio i the combiatio determies the ed behavior, just as the highest degree term did amog the algebraics. Vocabulary Ifiite Sequece Def: a sequece with a ifiite umber of terms. Coverget Sequece --Def: a ifiite sequece where a approaches a particular value whe goes to ifiity. Diverget Sequece --Def: a ifiite sequece where a does ot approach a particular value whe goes to ifiity. I some cases, a series of ifiite terms ca have a total, if the later terms get ifiitely small. Much of the series work i Calculus is about fidig whether a ifiite series is coverget (has a total) or diverget (has total keeps gettig bigger idefiitely). Geometric series are the easiest for which to fid a ifiite sum. I other words, 469

A sequece is coverget if ad oly if. lim a = c A sequece is diverget if ad oly if. lim a c Geometric ad Arithmetic series are the oes we ecoutered i Algebra 2. There are several other kids, some of which are importat for our study of series. We will look at them more closely later. Ex Which of the followig sequeces coverge? e 2 I. 4 II. III. 3 2 e 2 e 2 I each case, we just eed to check the Limit at Ifiity. That is, we eed to check the ed behavior. I. 4 lim ; therefore, this sequece coverges. 3 = 4 3 II. III. lim e2 lim ; therefore, this oe diverges. 2 = e 2 L'H e 2 = lim e2 2 e 2 2 = ; therefore, this coverges. So, oly I ad III coverge. This is a typical AP questio. The key is CRITICAL READING. Sometimes, the questio is about Sequeces, sometimes it is about Series (which we will cosider later it he chapter). Sometimes the questio is about covergece ad sometimes it is about divergece. 470

Ex 2 Which of the followig sequeces diverge? 3 I. 2 II. 7l III. 7 3 2 4 4 2 3 I. 3 2 ; therefore, this sequece does ot diverges. lim 7 3 = 0 II. lim 7l L'H 2 = lim 7 2 = 0; therefore, this oe does ot diverge. III. lim 4 4 ; therefore, this diverges. 2 3 = So, oly III diverges. Vocabulary Ifiite Sum--Def: the sum of all the terms of a sequece. This is ot possible for a arithmetic series, but might be for a geometric series. Coverget Series--Def: a ifiite series that has a total. Diverget Series--Def: a ifiite series that does ot have a total. I other words, A series is coverget if ad oly if. =c a = a = c A series is diverget if ad oly if. 47

Whe testig a umerical series for divergece or covergece, the two easiest tests are: Divergece Test (th term test): If lim a 0 a diverges. If = lim a = 0 o coclusio The Divergece Test basically says that if the terms at the ed are ot 0, the sum will just keep gettig larger ad larger (i.e., diverges). While easy to use, each has a disadvatage. The Divergece Test tells you if a fuctio has terms approachig 0. If the terms at the upper ed of the series do ot approach 0, the series caot coverge. But, there are may series with ed-terms approachig 0 that still diverge. So half the test is icoclusive. Ex 3 For which of the followig series is the Divergece Test icoclusive? I. 2 II. III. + = 2 lim 2 = 0 I. ; so the Divergece Test is icoclusive for I. Note that = this is ot eough to decide if it coverges or ot. = II. III. lim 2 + lim = 0 = 2 lim 2 + = ; therefore, II diverges. ; therefore, the Divergece Test is icoclusive for III. I ad III pass the Divergece Test 472

Kids of Series Arithmetic Series: a + = a + ( )d Geometric Series: a = a r [Coverget is r < ] p-series: [Coverget is p > ] = p Alteratig series: or ( cos π)a or ay other series where the a =0 sigs of the terms alterate betwee + ad -. Harmoic Series: =+ (Note that this is a p-series where p=) = 2 + 3 + 4 + [Diverget because p = ] Alteratig Harmoic Series: = = 2 + 3 4 + [Coverget by Alteratig Series Test] =0 + 473

A similar test to the Divergece Test, i process, is the Alteratig Series test: The Liebitz Alteratig Series Test If a series is Alteratig Series ad is a a +, the it is coverget if ad oly if lim a = 0. This meas the o-alteratig part of the series must be decreasig ad pass the Divergece Test (that is, the sequece coverges to 0). Ex 4 Is the Alteratig Harmoic Series coverget or diverget? Remember that the Alteratig Harmoic Series is. i) a = = ad ii) lim +, so > a + a + ( ) + = lim = 0 = + ( a + = ) +2 + = + So = ( ) + coverges. 474

Ex 5 Is = cos( π)! coverget or diverget? Remember that the Alteratig Harmoic Series is. cos π i) a = = ad!! ii), so! > a ( + a )! + ( ) + lim = lim = 0 a + = a = ( ) cos π + ( +)! = ( +)! So cos π! = coverges. 475

7.3 Homework Set A Determie if the sequece coverges or diverges. { }. 2. 3. 4. + L 2 5. 6. 3+ 2 2 + 2 2 3 + { 2 } 7.! ( + 2)! Determie which series might be coverget by the Divergece Test. 8. 9. = + 0. 3 8. = = 4 3 2 5 2 =0 2. =0 4 + 5 Test these series for covergece or divergece. 3. 4. = 3 + 4 2 + 5. ( ) 6. ( ) + L = = = 476

7. l ( ) 8. = 9. 20. 4 + = = = cos ( π ) 3 4 cos ( π ) 5 4 3 2. 22. = 7.3 Homework Set B Determie if the sequece coverges or diverges. = 2 4 +. 2 2 + 2. 3 5 { 2 + } { } 3. ta 2 3 4. 5. cos 6. e 3 { } 6 2 + 4 { } + 3 7. 8. 49 2 + + Determie which series might be coverget by the Divergece Test. 9. 5 4 0. 3 =0. 80 2 2. =0 2 = 3. 4. = =0 3 e 2 si 3 =0 ta ( 4) 477

7.4: The Itegral Test Whe testig a umerical series for divergece or covergece, oe of the easiest tests is: Itegral Test: a ad x dx either both coverge or both diverge. = The reaso the Itegral Test works is best explaied visually. The series ca be visualized as a sum of Riema Rectagles with width = ad height = the term values. If the itegral diverges (i.e., the sum is ifiite), the rectagles ca be draw left-had ad their sum is larger tha a ifiite umber. If the itegral coverges (i.e., the sum is fiite), the rectagles ca be draw right-had ad their sum is smaller tha a fiite umber. I both situatios, the fuctio must be decreasig. While easy to use, there is a disadvatage: The Itegral Test oly works o fuctios that ca be itegrated. Not all fuctios ca be itegrated. 478

Ex Use the Itegral Test to determie which of these series coverge. I. II. III. 2 + = I. ; therefore, I diverges. x dx = l x = de = =2 l 2 II. x 2 + dx = ta x = π 2 π 4 = π 4 ; therefore, II coverges. III. xl 2 x dx = 2 coverges. l 2 u du 2 = = 0 l2 u l 2 = l2 ; therefore, III II ad III coverge. Two otes about this example. First, I. proves that the Harmoic Series always diverges. Secod, III was a series that started at = 2 istead of =. This was to avoid the problem of the first term beig trasfiite. Ex 2 For what values of p does the p-series p = coverge? Lim p = 0 for all p>0, so p must be positive.. This itegral will be fiite as log as p>, because this x p dx = x p+ p + will leave x i the deomiator so the limit ca go to 0. Therefore, p = coverges for all p>. We will use this fact later i this chapter, whe applyig the Compariso Tests. 479

Ex 3 Determie whether 0 + 2 is coverget or diverget. 0 x + x 2 dx = lim b = 2 lim b b 0 x + x 2 dx x=b x=0 du u 0 b = 2 lim l + x2 b l = 2 lim l + b2 b = Sice this itegral diverges, the series 0 + 2 diverges. 480

7.4 Homework Set A Apply the Itegral Test to each of these series to determie divergece or covergece. If the series is coverget, state the improper itegral value.. = 2. 2 + 3 2 =2 l 2 3. 4. + 3 + = 5. + 6. 8 + 27 + 64 + 25 +... 5 2 = 3 = 4 7. 8. 2 + = = 2 + 9. 0. = 5 + 2 4 e 2 = 7.4 Homework Set B Apply the Itegral Test to each of these series to determie divergece or covergece. If the series is coverget, state the improper itegral value.. 2 e 3 2. = 3. 6 5 4. =+ 6 5. l 6. =2 3 2 =+ 6 =0 2 = 4 + 3 si 7. 8. =+ 2 = 4 48

7.5: The Direct ad Limit Compariso Tests The Direct ad Limit Compariso Tests are much less algebraic tha the Divergece ad Itegral Tests. They ivolve comparig a series to some series that we already kow coverges or diverges. So what do we kow? REMINDER: Factual Kowledge for Comparisos Geometric series:. a r coverges if r <ad diverges if r =0 2. If a r coverges, it coverges to =0 p-series:. coverges if p >ad diverges if p =0 p a 0 r Note Well: Harmoic Series: = + Diverges = 2 + 3 + 4 + Alteratig Harmoic Series: = + Coverges 2 + 3 + 4 + = 482

The Direct Compariso Test Whe comparig a give series to a kow series, b = b = ) if a b ad coverges, the coverges. 2) if a b ad diverges, the diverges. The use of this test is very verbal. a = a = a = b = The Limit Compariso Test a = b = Whe comparig a give series to a kow series,. If lim a = ay positive Real umber, the both coverge b or both diverge. 2. If lim a = 0, the a coverges if coverges. b = b = 3. If lim a = de, the a diverges if diverges. b = a = b = Key Idea I: What series to compare to depeds o where the variable is. If is i the base, compare to the p-series. If is i the expoet, compare to the Geometric Series. If is i both the base ad expoet (but ot ), try either. If, use the Nth Root Test (sectio 5-4). 483

Key Idea II: Which test to use depeds o the combiatio of the relative sizes ad whether the kow series coverges or diverges. b = coverges b = diverges a b Direct Compariso Test Limit Compariso Test a b Limit Compariso Test Direct Compariso Test Key Idea III: If a is ratioal, we wat to compare to the ed behavior model of the correspodig fuctio. OBJECTIVE Use the Compariso Tests to check for covergece or divergece. Ex Does = 3 + 2 coverge? We ca make a direct compariso betwee ad. 3 + 2 = = 3 because has a larger deomiator. Sice 3 + 2 < 3 3 + 2 = 3 coverges (it is a p-series with p>) ad is smaller tha, the 3 + 2 3 coverges. 3 + 2 = 484

Ex 2 Does =3 2 coverge? We ca make a direct compariso betwee ad. 2 = = because has a smaller deomiator. Sice diverges 2 > 2 = (it is the Harmoic Series--a p-series with p=) ad is bigger tha, 2 the diverges. 2 =3 Ex 3 Does = 2 4 coverge? If we ca try to make a direct compariso betwee ad, 2 4 = = 2 we fid that (because has a larger deomiator) but 2 4 > 2 2 4 coverges (it is a p-series with p>). So the Direct Compariso Test = 2 does ot apply. Lim 2 4 2 = Lim 2 2 4 = > 0 This limit is greater tha 0 ad coverges also. = 2 coverges. Therefore, = 2 4 485

Ex 4 Does = 2 coverge? because has a smaller deomiator. But 2 > 2 2 = 2 coverges (it is a Geometric Series with r<), so Direct Compariso will ot work. Lim 2 2 = Lim L 2 = ' H Lim 2 l2 2 l2 = 2 This limit is greater tha 0 ad coverges also. = 2 coverges. Therefore, = 2 Ex 5 Does = 2 + 3 coverge? We ca make a direct compariso betwee ad. 2 + 3 = = 2 because has a larger deomiator. Sice 2 + 3 < 2 2 + 3 = 2 coverges ad is smaller tha, the coverges. 2 + 3 2 = 2 + 3 486

Ex 6 Does = ( 3 + 2) 4 coverge? We ca use the Limit Compariso Test betwee ad. = 3 + 2 4 = 3 4 Lim ( 3 + 2) 4 3 4 = Lim 3 4 ( 3 + 2) 4 = = 3 4 diverges, therefore, = ( 3 + 2) 4 diverges. The oe formula from freshma year that will be useful is for the sum of a Ifiite Geometric Series: Sum of a Ifiite Geometric Series S = a r if ad oly if r <., If r, the series is diverget. 487

Ex 7 Fid the ifiite sum for the geometric sequece with a = 024 ad r =. 2 S = a r S = 024 2 = 682 2 3 Ex 8 Fid the ifiite sum for the geometric sequece with a = 24 ad r = 5. 4 Sice r = 5, the formula will yield a icorrect aswer. This series is 4 > diverget. 488

7.5 Homework Set A Test these series for covergece or divergece.. 2. 2 + + = 3. + 4. = 2 = = 5 2 + 3 3 2 5. 3 + 6. 3 = 7. 3+ cos 8. = 3 = = ( +)2 + 9. 2 5 0. si 3 + + = 7.5 Homework Set B Test these series for covergece or divergece. =. si 2 2. 3. = 3 + 3 =2 3 =! 4. 3 2 + 2 5 5. 6. = 4 2 2 + 7 =( + 2) 4 7. + 2si 8. 9. = 4 = + 4 = 4 + 2 ( + 3 ) 2 = 2 + 4 0. = 2 + 4 489

7.6: Absolute ad Coditioal Covergece The four tests i the previous two sectios oly apply to series of positive values. There is a separate test for alteratig series called the Liebitz Alteratig Series Test. The Alteratig Series Test (AST) If a series is Alteratig Series ad is a a +, the it is coverget if ad oly if Lim a = 0. This test was itroduced i Sectio 6.2. Vocabulary Absolute Covergece Def: Whe a alteratig series ad its absolute value are both coverget. Coditioal Covergece Def: Whe a alteratig series is coverget but its absolute value are diverget. OBJECTIVE Use the Alteratig Series Test to check for covergece or divergece. Ex Is = cos( π) 3 coverget or diverget? Remember that the Alteratig Harmoic Series is. cos π i) a = = ad a 3 3 + = a = ( ) = ( +) 3 ( +) 3 cos π + 490

ii) > 3 + Lim, so 3 a a + + 3 = Lim = 0 3 So = cos( π) 3 coverges. Ex 2 Is = cos( π) 3 absolutely or coditioally coverget? Ex proved that = cos( π) 3 is coverget by the AST. To demostrate coditioal vs. absolute covergece, we test the series without the alterator: cos π = = 3 = 3 is a p-series with p >, therefore, sice both the alteratig ad oalteratig versios of the series coverge, = cos π 3 = 3 coverges absolutely. 49

+ ( + 2) Ex 3 Is coverget or diverget? If coverget, is it absolutely or = ( +) coditioally coverget? i) a = ( + + 2) = + 2 ad + 2 + + 2, so 2 + > + 3 2 + 3 + 2 a + = + + ( +) + 2 ( +) + a a + = + 3 2 + 3 + 2 ii) Lim ( ) + + 2 + = Lim + 2 2 + = 0 So = ( ) + + 2 + coverges. To test for absolute vs. coditioal covergece, we eed to test + 2. + We ca do the Limit Compariso Test agaist the ed behavior model. = + 2 ( +) + 2 Lim = Lim ( +) = Lim 2 + 2 2 + = This limit is greater tha 0, therefore, both do the same thig--amely, they diverge. So, = = ( ) + + 2 + is coditioally coverget. 492

Ex 4 Is = ( ) + coverget or diverget? If coverget, is it absolutely or 2 coditioally coverget? + i) a = ad 2 = 2 a + = = 2 + 2 2, so 2 > a 2 2 a + + +2 + Lim = ii) 2 Lim 2 = 0 So coverges. Furthermore, coverges because it is a Geometric series with r =. So, 2 2 ( ) + coverges absolutely. = 2 = 2 493

7.6 Homework Set A Test these series for covergece or divergece. If it is coverget, determie if it has absolute or coditioal covergece.. 2. = 2 4 + 3. ( ) 4. ( ) l = + 5. ( ) l 6. cosπ = = = = 3 4 ( ) + 7. 8. 4 2 + = = ( )! 7.6 Homework Set B a) Test these series for covergece or divergece. b) If it is coverget, determie if it has absolute or coditioal covergece. + 3. 2. 3. = 2 4 ( + 5) =2 4. 5. 6. + + = 4 = = = ( ) + cos( π)! si π 2 7. 8. 9. 6 = = = ( ) arcta cos π 0.. 2. = arcta π = e 2 = si π 2 494

7.7: The Ratio ad Nth Root Tests The Ratio Test is oe of the more easily used tests ad it works o most series, especially those that ivolve factorials ad those that are a combiatio of geometric ad p-series. Ufortuately, it is sometimes icoclusive, especially with alteratig series. Cauchy Ratio Test: If lim a + < it coverges; a If If lim a + > a lim a + = a it diverges; the test is icoclusive. The Nth Root Test is best for series with is both the base ad the expoet. The Nth Root Test: If lim a < it coverges; If If lim a > it diverges; lim a = the test is icoclusive. OBJECTIVE Use the Ratio ad Nth Root Tests to check for covergece or divergece. Sice the Cauchy Ratio Test ivolves Limits at Ifiity, It might be a good idea to review the Hierarchy of fuctios: 495

The Hierarchy of Fuctios. Logs grow the slowest. 2. Polyomials, Ratioals ad Radicals grow faster tha logs ad the degree of the EBM determies which algebraic fuctio grows fastest. For example, y = x 2grows more slowly tha y = x 2. 3. The trig iverses fall i betwee the algebraic fuctios at the value of their respective horizotal asymptotes. 4. Expoetial fuctios grow faster tha the others. (I BC Calculus, we will see the factorial fuctio, y =!, grows the fastest.) 5. The fastest growig fuctio i the combiatio determies the ed behavior, just as the highest degree term did amog the algebraics. 496

Ex Does =! coverge? lim a + a = lim ( +)!! = lim! ( +)! = lim = 0! ( +)! = lim + Sice this limit is <, =! coverges. Ex 2 Does = 2 coverge? Lim a + a = Lim + 2 + 2 = Lim ( +)2 2 + = Lim + 2 = 2 Sice this limit is <, = 2 coverges. 497

Ex 3 Does = + 2 + coverge? Lim a + a = Lim = Lim ( +) + 2 + ( +) + + 2 ( +) + 3 ( + ) ( + 2) ( + ) + 2 2 + 3 = Lim 2 + 4 + 4 = So the Ratio Test is icoclusive. The easiest test would be the Limit Compariso Test: + 2 Compare to. = ( +) = + 2 ( +) + 2) lim = lim ( +) = Sice the Limit =,both series coverge or both diverge. therefore, = diverges, = + 2 + diverges also. Ex 4 Does = l + 2 coverge? This series has to a power; therefore, the Root Test is appropriate. 498

Lim l = ( + 2) Lim Sice this limit is <, = l( + 2) = 0 l + 2 coverges. Ex 5 Does = + e coverge? This series has to a power; therefore, the Root Test is appropriate. Lim + e = Lim Sice this limit is >, + e = Lim ( + )e = e > = + e diverges. 499

7.7 Homework Set A Test these series for covergece or divergece.. 2 2. 3. 2 3 = = = 4. ( ) 5. ( ) 6.! 5 + = = = ( 2)! cos π 7. 3 8.! = = 0 ( +)4 2+ 2 + 9. 0. 2 2 + = = 3 +3. For which of the followig series is the Ratio Test icoclusive. What test would you use istead? (a) (b) 3 (c) (d) 3 2 = = = = + 2 7.7 Homework Set B Test these series for covergece or divergece. 4. 2. 3. e! = 4 4 =2 4 = 3 5 7... 2 + 4. 5.! = l =2 6. 3 7. 3 8. =0 =0 =! 4 7 0... 3 + 500

4 9. 0. 2 3. + 5! = = = e 2! 2. = ( 3) 3 3. For which of the followig series is the Ratio Test icoclusive? What test would you use istead? (a) 2 (b) (c) (d) = 2 + 2 =( 3 )! =2 4 2 + 2 =3 3+ 50

7.8 Geeral Series Summary A series is coverget if ad oly if. a = a = = c c A series is diverget if ad oly if. Previously, we ivestigated the seve tests that will help determie if a give series coverges or ot. Those tests were:. Divergece Test (th term test): If Lim a 0 it diverges. If Lim a = 0 o coclusio 2. Itegral Test: If f(x) is a decreasig fuctio, the a ad x dx either both coverge or both diverge. = 3. Cauchy Ratio Test: If Lim a + < it coverges; a If Lim a + > a it diverges; If Lim a + a = the test is icoclusive. 4. The Alteratig Series Test: A Alteratig Series is coverget if ad oly if i) a a +, ad ii) Lim a = 0. 502

5. The Nth Root Test: If Lim a < it coverges; If If Lim a > it diverges; Lim a = the test is icoclusive. 6. The Direct Compariso Test: Whe comparig a give series to a kow series, i) if a b ad coverges, the coverges. ii) if a b ad diverges, the diverges. 7. The Limit Compariso Test: Whe comparig a give series to a b = kow series, b = b = b = a = a = a = a = i. If Lim a > 0the both coverge or both diverge. b ii. If Lim a = 0, the a coverges if coverges. b = b = iii. If Lim a =, the a diverges if diverges. b = b = OBJECTIVE Fid whether a give umerical series coverges or diverges. 503

504

7.8 Homework ) Which of the followig sequeces coverge? 3 I. 5 cos II. III. 7 4 π! e 2 (A) I oly (B) II oly (C) I ad III oly (D) II ad III oly (E) III oly 2) Which of the followig sequeces diverge? l 5cos3 I. II. III. 5 e! ( + 2 )! (A) II oly (B) III oly (C) I ad II oly (D) I ad III oly (E) I, II, ad III 3) Which of the followig series diverge? I. 9 8 + l5 II. III. cos 2π = (A) I oly (B) III oly (C) I ad II oly =3 =2 2 3 + ( 2e) (D) II ad III oly (E) I, II, ad III 4) What are all values of k for which the ifiite series k coverges? 5 (A) k < 5 (B) k 5 (C) k < (D) k (E) k = 0 = 505

= si x 5) If f x 2, the =0 f π 6 = 4 3 (A) 2 (B) (C) (D) (E) diverget 3 2 2 6) Which of the followig series coverge? I. 2 + 3 II. + III. = 2 + 3 + 6 = e 2 =+ 4 (A) I oly (B) III oly (C) I ad III oly (D) II ad III oly (E) I ad II oly 7) Which of the followig series are diverget? e I. II. III. =0 3 + =0 ( + 5)!! =3 (A) I oly (B) II oly (C) III oly (D) I ad II oly (E) II ad III oly 8) Which of the followig series are coditioally coverget? e I. II. III. =0 ( + 5)! =0 = cos( π) (A) I oly (B) II oly (C) III oly (D) I ad II oly (E) I ad III oly 506

9) Which of the followig series are diverget? I. ( ) ( + 2)! II. III. 3 +! =0 =0! =3 (A) I ad III oly (B) II oly (C) III oly (D) I ad II oly (E) II ad III oly 0) What are all values of p for which the ifiite series diverges? (A) (B) (C) (D) (E) p > 2 p 2 p < 2 p 2 All real values of p 5 + p = 507

Numerical Series Test Calculator Allowed. L I. II. III. L 2 a = ( ) 3 3 3 + 2 2 + (A) I oly (B) III oly (C) I ad II oly (D) I ad III oly (E) I, II, ad III 2. Which of the followig Series diverge? I. 5 6 3 II. 5 III. = = 74 =5 (A) I oly (B) III oly (C) I ad II oly (D) II ad III oly (E) I, II, ad III 3. Which of the followig are series diverget? I. e l5 2 II. III. =( 3 + + 5)! =3cos( 2π) =2 ( 2e) (A) I oly (B) III oly (C) I ad III oly (D) II ad III oly (E) I, II, ad III 508

4. What are all values of p for which the ifiite series coverges? 5 + p = (A) p > 2 (B) p 2 (C) p < 2 (D) p 2 (E) All real values of p 5. Which of the followig are series coverget? I. 5 II. III. = 3 + 3 =4 + ( 3 + 7) 2 =2 (A) I oly (B) III oly (C) I ad II oly (D) I ad III oly (E) II ad III oly 6. Which of the followig series diverge? I. II. III. 2 + = = cos 3 = ( ) + (A) I oly (B) II oly (C) I ad II oly (D) I ad III oly (E) I, II ad III 509

a = 7. If coverges, the which of the followig is true? I. diverges II. III. a = a = = 4a coverges absolutely coverges (A) I oly (B) II oly (C) III oly (D) II ad III oly (E) I, II ad III 8. Which of the followig series are coditioally coverget? I. ( ) ( ) ( 2) II. III. L ta = 2 =2 = (A) I oly (B) I ad III oly (C) II ad III oly (D) I ad III oly (E) I, II, ad III 9. si π 6 = = (A) (B) 2 (C) 3 3 (D) 2 (E) diverget 2 3 2 0. Which of the followig three tests will establish that the series coverges? = 2 5 + 50

I. Direct Compariso Test with II. Limit Compariso Test with III. Direct Compariso Test with = 5 2 (A) I oly (B) II oly (C) I ad II oly (D) II ad III oly (E) I, II, ad III = = 3 2 2 Numerical Series Test Calculator Allowed. Use the Itegral Test to determie if 2 e 3 is coverget or diverget. = 2. Use the Ratio Test to determie if 2 3 is coverget or diverget.! = 3. Determie whether the series coverges or diverges. =( + 4) 2 5

7. Homework. 35 2. -382 3. -70 4. 5 5. 236 6. 7. 2 is ot i the arithmetic sequece. 3 8. 7 th 9. 49, 56, 63 0. 03 2, 00, 96 2, 93, 89 2. 38 5 7, 30 3 7, 22 7, 36 7, 5 4 7, 2 5 7 2. arithmetic; d = 5 3. geometric; r = 3 r = 2 4. geometric; 5. either 6. either 7. geometric; r = 5 6 8. 354294 9. 2.85 x 0 4 20..90603029 2. 32 22. 000 243 23. -656 24. 0 25. 7th 256 26. 5 ad 45 27. 28. 05, 2, 5 2 5 3, 5, 5 3 52

7.2 Homework. 37 60 2. 0 3. -7 4. 248 5..243 6. 45 7. 25 8. -57.5 9. -43 0. 37. 43 2. 4 3. 4 4. 222 5. -4040 6. 53,440 7. -3.843 x 0 7 8. = 6, S = -3070 9. 25 20. 5.6 2. Diverget 7.3 Homework. Diverget 2. Coverget 3. Coverget 4. Coverget 5. Coverget 6. Coverget 7. Coverget 8. Might Coverge 9. Might Coverge 0. Diverget. Diverget 2. Might Coverge 3. Coverget 4. Coverget 5. Coverget 6. Diverget 7. Coverget 8. Coverget 9. Coverget 20. Coverget 2. Diverget 22. Diverget 7.4 Homework. Coverget, 2. Coverget, 3. Diverget l2 2 53

4. Coverget, 5. Coverget, 6. Coverget, 7 2 6 2 7. Coverget, π 8. Diverget 9. Coverget,.0002 4 0. Coverget, 2e 7.5 Homework. Coverges by LCT 2. Coverges by LCT 3. Diverges by LCT 4. Coverges by the Ratio Test 5. Diverges by the Divergece Test 6. Coverges by the Ratio Test 7. Coverges by DCT 8. Diverges by LCT 9. Diverges by LCT 0. Diverges by LCT 7.6 Homework. Coditioally Coverget 2. Diverges 3. Coditioally Coverget 4. Diverges 5. Coditioally Coverget 6. Coditioally Cov. 7. Absolutely Coverget 8. Diverget 7.7 Homework. Diverget by Divergece Test 2. Coverges by the Itegral Test 3. Diverget by AST 4. Coverges by AST 5. Diverget by Divergece Test 6. Coverges by Ratio Test 7. Coverges by Ratio Test 8. Coverges by Ratio Test 9. Coverges by Root Test 0. Diverges by Root Test 54

a. Ratio Test icoclusive b. Ratio Test works c. Ratio Test works d. Ratio Test icoclusive 7.8 Homework. B 2. A 3. D 4. A 5. B 6. D 7. E 8. C 9. E 0. B 55